What are the rings whose countable power has the property that every finitely generated submodule is free?

1$\begingroup$ Are your rings commutative with unity? $\endgroup$– Alex KruckmanCommented Jun 12 at 23:05

$\begingroup$ @Alex Kruckman I would start with a commutative rings with unity. Once that becomes clarified, one can look at the more general case(s) $\endgroup$– RadoCommented Jun 15 at 3:46
2 Answers
I think your condition is equivalent to being a semifir (but see the remark below). See Chapter 2 of Paul Cohn, "Free ideal rings and localization in general rings". A commutative semifir is a Bézout domain, so in the commutative case your condition is equivalent to being a Bézout domain. This is more general than a PID.
For a semifir, you also need the ranks of finitely generated free modules to be well defined, so your condition might be a bit more general. Of course, in the commutative case, this is automatic.

$\begingroup$ Great! Interestingly (or strangely), I have never seen this book of Cohn. Worth checking it out. Some people use tensoring to define rank, Another way is to imitate vector spaces. $\endgroup$– RadoCommented Jun 16 at 22:12

$\begingroup$ Thanks. I will check Cohn's book, not sure I have it. $\endgroup$– RadoCommented Jun 16 at 22:24
Assuming rings are commutative with unity, these are exactly the Bézout domains (and the zero ring).
Let $R$ be a nonzero ring. The following are equivalent:
 $R$ is a Bézout domain (a domain in which every f.g. ideal is principal).
 Every f.g. torsionfree module is free.
 Every f.g. submodule of $R^\mathbb{N}$ is free.
 Every f.g. ideal is a free module.
$1\implies 2$ is the nontrivial implication. See Pete Clark's answer here. It also follows from Lemma 15.22.7 and Lemma 15.124.8 on the Stacks project.
$2\implies 3$ since $R^\mathbb{N}$ is torsionfree (so every submodule is torsionfree).
$3\implies 4$ since every ideal is a submodule of $R$, and hence a submodule of $R^\mathbb{N}$.
For $4\implies 1$, suppose $R$ is not a Bézout domain. Then $R$ has zerodivisors or a f.g. nonprincipal ideal. If $a\in R$ is a zerodivisor, then $(a)$ is a finitely generated ideal which is not a free module because it has torsion. If $I\subseteq R$ is a f.g. nonprincipal ideal, then $I$ is not a free module. Indeed, if $a,b\in I$ are nonzero, then $baab=0$, so $I$ has rank at most $1$, but it is not principal, hence not free.
As you can see, there is nothing at all special about $R^{\mathbb{N}}$ being used in the proof. You can replace $R^{\mathbb{N}}$ with any other nonzero torsionfree module, and the equivalence holds.


$\begingroup$ Interesting, we can then simply require the condition to be applied to R. $\endgroup$– RadoCommented Jun 16 at 22:26